## Re: [EMHL] Unique triangle

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• Dear Barukh, The answer is yes. Let BD, CE be the bisectors and I the incenter of triangle ABC. The circle (c1) (B, BD) meets BC at D at right of B and the
Message 1 of 2 , Nov 1 12:36 PM
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Dear Barukh,

Let BD, CE be the bisectors
and I the incenter of triangle ABC.
The circle (c1) (B, BD) meets BC at D'
at right of B and
the circle (c2) (C, CE) meets BC at E'
at left of C.
Since BD + CE > BI + CI > BC the point
D' is at right of E'.
When the angle B increases from 0 to 180
in order to construct ABC with BC = a and BD = t_b
the point D moves on the circle (c1) we double
the angle CBD and with line CD we find the vertex A.
The locus of A is a curve (c3) that passes through D'.
The orthogonal projection A' of A on BC moves
from D' to the left as angle B increases.
Similarly in order to construct ABC
with BC = a and CE = t_c the point E moves on
the circle (c2), the point A moves on
a curve (c4) that passes through E' and
the point A' moves from E' to the right.
Hence the curves (c3) and (c4) have only one
point of intersection (on the positive semiplane
relative to BC) and the triangle ABC
with BC = a, BD = t_b, CE = t_c is unique.

Best regards

>Dear Hyacinthians,
>
>Is it true that there exists *at most* one triangle with
>given elements a, t_b, t_c (the last two are angle
bisectors)?
>
>If yes, how can we prove this uniqueness?
>
>Thanks,
>Barukh.
>
>
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• Dear Nikos, Thank you very much for the explanations. Barukh.
Message 2 of 2 , Nov 2 11:10 PM
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Dear Nikos,

Thank you very much for the explanations.

Barukh.

> Dear Barukh,
>
>
> Let BD, CE be the bisectors
> and I the incenter of triangle ABC.
> The circle (c1) (B, BD) meets BC at D'
> at right of B and
> the circle (c2) (C, CE) meets BC at E'
> at left of C.
> Since BD + CE > BI + CI > BC the point
> D' is at right of E'.
> When the angle B increases from 0 to 180
> in order to construct ABC with BC = a and BD = t_b
> the point D moves on the circle (c1) we double
> the angle CBD and with line CD we find the vertex A.
> The locus of A is a curve (c3) that passes through D'.
> The orthogonal projection A' of A on BC moves
> from D' to the left as angle B increases.
> Similarly in order to construct ABC
> with BC = a and CE = t_c the point E moves on
> the circle (c2), the point A moves on
> a curve (c4) that passes through E' and
> the point A' moves from E' to the right.
> Hence the curves (c3) and (c4) have only one
> point of intersection (on the positive semiplane
> relative to BC) and the triangle ABC
> with BC = a, BD = t_b, CE = t_c is unique.
>
> Best regards
>
> >Dear Hyacinthians,
> >
> >Is it true that there exists *at most* one triangle with
> >given elements a, t_b, t_c (the last two are angle
> bisectors)?
> >
> >If yes, how can we prove this uniqueness?
> >
> >Thanks,
> >Barukh.
> >
> >
> >------------------------ Yahoo! Groups